「BZOJ1689」[Usaco2005 Open] Muddy roads 泥泞的路
Description
Farmer John has a problem: the dirt road from his farm to town has suffered in the recent rainstorms and now contains (1 <= N <= 10,000) mud pools. Farmer John has a collection of wooden planks of length L that he can use to bridge these mud pools. He can overlap planks and the ends do not need to be anchored on the ground. However, he must cover each pool completely. Given the mud pools, help FJ figure out the minimum number of planks he needs in order to completely cover all the mud pools.
Input
* Line 1: Two space-separated integers: N and L * Lines 2..N+1: Line i+1 contains two space-separated integers: s_i and e_i (0 <= s_i < e_i <= 1,000,000,000) that specify the start and end points of a mud pool along the road. The mud pools will not overlap. These numbers specify points, so a mud pool from 35 to 39 can be covered by a single board of length 4. Mud pools at (3,6) and (6,9) are not considered to overlap.
Output
* Line 1: The miminum number of planks FJ needs to use.
Sample Input
1 6
13 17
8 12
Sample Output
HINT
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题解
考试期间刷刷水还是很高兴的。。
按s排序然后贪心，木板尽量后放
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#include<iostream> #include<cstdio> #include<cstring> #include<algorithm> using namespace std; int n,l,now,ans; struct data{int x,y;}a[10001]; bool cmp(data a,data b) {return a.x<b.x;} int main() { scanf("%d%d",&n,&l); for(int i=1;i<=n;i++) scanf("%d%d",&a[i].x,&a[i].y); sort(a+1,a+n+1,cmp); for(int i=1;i<=n;i++) { if(now>=a[i].y)continue; now=max(now,a[i].x); while(now<a[i].y){ans++;now+=l;} } printf("%d",ans); return 0; } |
直接就能A了，不过我们可以优化一下，计算每个坑要放的木板一下加上去
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#include<iostream> #include<cstdio> #include<cstring> #include<algorithm> using namespace std; int n,l,now,ans; struct data{int x,y;}a[10001]; bool cmp(data a,data b) {return a.x<b.x;} int main() { scanf("%d%d",&n,&l); for(int i=1;i<=n;i++) scanf("%d%d",&a[i].x,&a[i].y); sort(a+1,a+n+1,cmp); for(int i=1;i<=n;i++) { if(now>=a[i].y)continue; now=max(now,a[i].x); int t; t=(a[i].y-now)/l; if((a[i].y-now)%l)t++; ans+=t;now+=t*l; } printf("%d",ans); return 0; } |
OI 课件=>
